Problem: Suppose we have a plane $P$ defined by the transformation $T$ for $-1 < u < 1$ and $0 < v < 1$. $T(u, v) = (4u - 6v, -u + 2v, v)$ What is the surface area of $P$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $12\sqrt{2}$ (Choice B) B $8\sqrt{15}$ (Choice C) C $8\sqrt{2}$ (Choice D) D $2\sqrt{21}$
Assume we have a surface $S$ parameterized by a transformation $T$. If we want to find the surface integral over $S$ of a function $f$, we can use the formula below to convert it into a familiar double integral. $ \iint_S f(T(u, v)) | T_u \times T_v | \, du \, dv$ Finding surface area using a surface integral means using $f(x, y, z) = 1$. In effect, we are saying that we only care about the scaling factor caused by the area element. Therefore: $A = \int_{-1}^{1} \int_{0}^1 |T_u \times T_v| \, du \, dv$ Now we need to find the magnitude of the area element. $|T_u \times T_v| = \sqrt{21}$ [Calculation] The final step is to evaluate the double integral. $\begin{aligned} A &= \int_{-1}^{1} \int_{0}^1 |T_u \times T_v| \, du \, dv \\ \\ &= \int_{-1}^1\int_0^1 \sqrt{21} \, du \, dv \\ \\ &= \sqrt{21} \int_{-1}^1 \, dv \\ \\ &= 2 \sqrt{21} \end{aligned}$ In conclusion, the surface area of $P$ is $2\sqrt{21}$.